Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12



  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Ex 6.1, 15 The total cost C(๐‘ฅ) in Rupees associated with the production of ๐‘ฅ units of an item is given by ๐ถ(๐‘ฅ) = 0.007๐‘ฅ^3 โ€“ 0.003๐‘ฅ2 + 15๐‘ฅ + 4000. Find the marginal cost when 17 units are produced. Since Marginal Cost is Rate of change in Total Cost w.r.t No of units produced Let MC be the marginal cost & C (๐‘ฅ) is the total cost & ๐‘ฅ be the number of units produced So, MC = ๐‘‘(๐ถ(๐‘ฅ))/๐‘‘๐‘ฅ โ€ฆ(1) It is Given that total cost C(๐‘ฅ)=0.007๐‘ฅ^3โˆ’0.003๐‘ฅ^2+15๐‘ฅ+4000 & we need to find Marginal cost when 17 units produced i.e. MC at ๐‘ฅ=17 Now MC = ๐‘‘(๐ถ(๐‘ฅ))/๐‘ฅ MC = ๐‘‘(0.007๐‘ฅ^3 โˆ’ 0.003๐‘ฅ^2 + 15๐‘ฅ + 4000)/๐‘‘๐‘ฅ MC = ๐‘‘(0.007๐‘ฅ^3 )/๐‘‘๐‘ฅ โˆ’ ๐‘‘(0.003๐‘ฅ^2 )/๐‘‘๐‘ฅ+ ๐‘‘(15๐‘ฅ)/๐‘‘๐‘ฅ+ ๐‘‘(4000)/๐‘‘๐‘ฅ MC = 0.007 ๐‘‘(๐‘ฅ^3 )/๐‘‘๐‘ฅ โˆ’0.003 ๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ +15๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ+0 MC = 0.007 ร— 3๐‘ฅ^2โˆ’0.003 ร—2๐‘ฅ+15 MC = 0.021๐‘ฅ^2โˆ’0.006๐‘ฅ+15 We need to find MC when ๐‘ฅ=17 Putting ๐‘ฅ=17 MC = 0.021(17)^2โˆ’0.006(17)+15 MC = 6.069 โ€“ 0.102 + 15 MC = 20.967 Hence the Required Marginal cost is Rs. 20.967 when ๐‘ฅ=17

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.